Help with sine and cosine integral

[yungman]

I am verifying the equation of radiation power of dipole antenna. I found mistakes in the derivation in the notes. I know the final equation is correct. So instead of following the steps in the notes, I reverse the step by using the final formula and going back step by step. Here is the final equation, $$P_{rad}=η\frac {|I_0|^2}{4\pi}I_{int}\;\hbox{ where:}$$
$$I_{int}=\left[ C +ln(kl)-Ci(kl)+\frac 1 2 \sin(kl)[Si(2kl)-2Si(kl)]+\frac 1 2 \cos(kl)\left(C+ln(\frac{kl}{2})+Ci(2kl)-2Ci(kl)\right)\right]$$
Where $C$ is Euler constant and $C$=0.5772.

The step before this that I want to verify is:
$$I_{int}=[1+\cos(kl)]\int_0^{kl}\frac{1-cos(v)}{v}dv-2sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\sin(kl)\int_0^{2kl}\frac{\sin(v)}{v}dv-\cos(kl)\int_0^{2kl}\frac{1-\cos(v)}{v}dv$$

I know
$$Si(kl)=\int_0^{kl}\frac{\sin(v)}{v}dv,\;Si(2kl)=\int_0^{2kl}\frac{\sin(v)}{v}dv,\;<br /> Ci(kl)=-\int_0^{kl}\frac{1-\cos(v)}{v}dv+log(kl)+C,\;Ci(2kl)=\int_0^{2kl}\frac{1-\cos(v)}{v}dv+log(2kl)+C$$

I have not idea where $ln(kl), \;ln(\frac{kl}{2})$ come from. How come it is ln instead of log? Also, it will make more sense if it is $ln(2kl)$ rather than $ln(\frac{kl}{2})$. Please help.

Thanks

 

[SteamKing]

Just a comment on your Latex: in the expressions Iint =, Si (kl) = and Ci (kl) = , the 'i' looks like a subscript.

 

[yungman]

Just a comment on your Latex: in the expressions Iint =, Si (kl) = and Ci (kl) = , the 'i' looks like a subscript.

I thought it's a subscript and I did use subscript.

 

[SteamKing]

Si and Ci (without the 'i' being a subscript) are used to denote the Sine Integral and the Cosine Integral, respectively. I know it could be confusing, but these are the accepted symbols for these functions. I wouldn't want someone to go looking for a subscript where none was present.

 

[yungman]

Si and Ci (without the 'i' being a subscript) are used to denote the Sine Integral and the Cosine Integral, respectively. I know it could be confusing, but these are the accepted symbols for these functions. I wouldn't want someone to go looking for a subscript where none was present.

Ok, I went back and change that already. I modified a little as I was wrong on the Ci

$$Ci(kl)=-\int_0^{kl} \frac{1-\cos(v)}{v}dv+log(kl)+C$$

 

[yungman]

$$Si(kl)=\int_0^{kl}\frac{\sin(v)}{v}dv,\;Si(2kl)=\int_0^{2kl}\frac{\sin(v)}{v}dv,\;<br /> Ci(kl)=-\int_0^{kl}\frac{1-\cos(v)}{v}dv+log(kl)+C,\;Ci(2kl)=\int_0^{2kl}\frac{1-\cos(v)}{v}dv+log(2kl)+C$$

I am trying to work on it and this is what I have:
$$I_{int}=\left[ C +ln(kl)-Ci(kl)+\frac 1 2 \sin(kl)[Si(2kl)-2Si(kl)]+\frac 1 2 \cos(kl)\left(C+ln(\frac{kl}{2})+Ci(2kl)-2Ci(kl)\right)\right]$$
Where $C$ is Euler constant and $C$=0.5772.
$$\Rightarrow\;I_{int}=\int_0^{kl}\frac{1-\cos(v)}{v}dv+\frac{\sin(kl)}{2}\int_0^{2kl}\frac{s\sin(v)}{v}dv-\sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\frac{\cos(kl)}{2}\left( C+ln(\frac {kl}{2})+Ci(2kl)-2Ci(kl)\right)$$
Where I cannot resolve $\frac{\cos(kl)}{2}\left( C+ln(\frac {kl}{2})+Ci(2kl)-2Ci(kl)\right)$The step in the notes I want to verify is:
$$I_{int}=[1+\cos(kl)]\int_0^{kl}\frac{1-cos(v)}{v}dv-2sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\sin(kl)\int_0^{2kl}\frac{\sin(v)}{v}dv-\cos(kl)\int_0^{2kl}\frac{1-\cos(v)}{v}dv$$

It obviously doesn't match. Attacked is the scanned original notes:

 

[jackmell]

I doubt seriously that's log base ten. They just used "log" to represent natural log as we do in Complex Analysis, that is, the expression log(u) in Complex Analysis means the natural log base e and I bet that's what is meant in your expression.

 

[yungman]

I doubt seriously that's log base ten. They just used "log" to represent natural log as we do in Complex Analysis, that is, the expression log(u) in Complex Analysis means the natural log base e and I bet that's what is meant in your expression.

I kind of suspected that, I just want to verify.

I just posted my work right before your post, can you take a look at it?

Thanks

 

[jackmell]

Ok. Mathematica does not give your answer. Rather it gives:

$$\gamma-\text{Ci}(v)-\cos(v) \text{Ci}(v)+\cos(v)\text{Ci}(2v)-\cos(v)\log(2)+\log(v)-2\sin(v)\text{Si}(v)+\sin(v)\text{Si}(2v)$$

where v =kl and gamma is Eulergamma. Now, there's always the chance that they are two forms of the same answer. Easy to check that. Just plug in some real number for kl and see if your answer agrees with Mathematica. If your doesn't, no offense but Mathematica is pretty sharp and I'd have to go with it and assume you made a mistake. Don't have time to work through it now though.

 

[yungman]

Ok. Mathematica does not give your answer. Rather it gives:

$$\gamma-\text{Ci}(v)-\cos(v) \text{Ci}(v)+\cos(v)\text{Ci}(2v)-\cos(v)\log(2)+\log(v)-2\sin(v)\text{Si}(v)+\sin(v)\text{Si}(2v)$$

where v =kl and gamma is Eulergamma. Now, there's always the chance that they are two forms of the same answer. Easy to check that. Just plug in some real number for kl and see if your answer agrees with Mathematica. If your doesn't, no offense but Mathematica is pretty sharp and I'd have to go with it and assume you made a mistake. Don't have time to work through it now though.

Thanks for the help, did you use (5) in the scanned notes to derive this? I don't trust the notes, It was proven from the other posts that there are mistakes. When I worked from the other end( the beginning), I don't get the $2\sin(v)\text{Si}(v)$. There should not be a "2" in it. That's the reason I go from the end equation back this time as I am not going anywhere going from the start to derive the equation.

I double checked with Wikipedia already that the end equation is correct for sure. So that is my starting point to derive back to the electromagnetic wave equations. If your result of the Mathematica from (5) doesn't match (6). There must be a problem with (5) of the scanned notes.

That's the problem with those graduate level books, they are not as well scrutinized and there can be typos and errors. You can't take anything for granted and have to verify every single equation. This is supposed to be the real solution manual already! And here I am, struggling with my rusty math!

let me attach the complete notes of this section. This is page 1, the attachment in the early post is the continuation in page 2. It is proven there is a mistake from (1) to (2) already. And it does not add up from (2) to (3)!

 

[yungman]

Anyone please?

 

[yungman]

Ok. Mathematica does not give your answer. Rather it gives:

$$\gamma-\text{Ci}(v)-\cos(v) \text{Ci}(v)+\cos(v)\text{Ci}(2v)-\cos(v)\log(2)+\log(v)-2\sin(v)\text{Si}(v)+\sin(v)\text{Si}(2v)$$

where v =kl and gamma is Eulergamma. Now, there's always the chance that they are two forms of the same answer. Easy to check that. Just plug in some real number for kl and see if your answer agrees with Mathematica. If your doesn't, no offense but Mathematica is pretty sharp and I'd have to go with it and assume you made a mistake. Don't have time to work through it now though.

I work out the equation:
$$I_{int}=[1+\cos(kl)]\int_0^{kl}\frac{1-cos(v)}{v}dv-2sin(kl)\int_0^{kl}\frac{\sin(v)}{v}dv+\sin(kl)\int_0^{2kl}\frac{\sin(v)}{v}dv-\cos(kl)\int_0^{2kl}\frac{1-\cos(v)}{v}dv$$
$$\Rightarrow\; I_{int}=-Ci(kl)+ln(kl)+C-\cos(kl)Ci(kl)+\cos(kl)ln(kl)-2\sin(kl)Si(kl)+\sin(kl)Si(2kl)+\cos(kl)Ci(2kl)-\cos(kl)ln(2kl)$$

The difference from yours is I have $\cos(kl)ln(kl)$ you don't have. Also you have $-\cos (kl) ln(2)$. I have $-\cos(kl)ln(2kl)$. Are you sure the $ln(2)$ is not $ln(2kl)$?

Can you double check as I don't have Mathematica. If it is online calculator, please give me the link to it.

Thanks

 

[yungman]

Anyone, please?

 

[Bill Simpson]

Subtracting your two equations and simplifying

FullSimplify[
(1 + Cos[kl]) Integrate[(1 - Cos[v])/v, {v, 0, kl}] - 2 Sin[kl] Integrate[Sin[v]/v, {v, 0, kl}] +
Sin[kl] Integrate[Sin[v]/v, {v, 0, 2 kl}] - Cos[kl] Integrate[(1 - Cos[v])/v, {v, 0, 2 kl}]
-
(-CosIntegral[kl] + Log[kl] + c - Cos[kl] CosIntegral[kl] + Cos[kl] Log[kl] - 2 Sin[kl] SinIntegral[kl] + Sin[kl] SinIntegral[2 kl] + Cos[kl] CosIntegral[2 kl] - Cos[kl] Log[2 kl])
]

gives

-c + EulerGamma

so if your C equals EulerGamma then the two are the same.

Note: I've changed the case of characters to satisfy Mathematica.

 

[yungman]

Thanks for the help. I got it. I was so rusty that I forgot the trick of the log function. It just dawn on me late last night

$$2ln(kl)-ln(2kl)=ln\left(\frac {k^2l^2}{2kl}\right)=\ln\left(\frac{kl}{2}\right)$$
$$ln(kl)-ln(2kl)=-ln(2)$$

That proofs the equations. Thanks for the help from everyone here.